Answer
$\frac{\pi}{6}$
Work Step by Step
$$g'(x)=(\arcsin(\frac{x}{4})+x(\arcsin(\frac{x}{4}))')+\frac{-2x}{2\sqrt{16-x^2}}$$
$$g'(x)=(\arcsin(\frac{x}{4})+x(\frac{x}{4})'(\arcsin)'(\frac{x}{4}))+\frac{-2x}{2\sqrt{16-x^2}}$$
$$g'(x)=(\arcsin(\frac{x}{4})+\frac{x}{4}\frac{1}{\sqrt{1-(\frac{x}{4})^{2}}})+\frac{-2x}{2\sqrt{16-x^2}}$$
$$g'(x)=(\arcsin(\frac{x}{4})+\frac{x}{4}\frac{1}{\sqrt{1-(\frac{x}{4})^{2}}})-\frac{x}{\sqrt{16-x^2}}$$
so:
$$g'(2)=(\arcsin(\frac{2}{4})+\frac{2}{4}\frac{1}{\sqrt{1-(\frac{2}{4})^{2}}})-\frac{2}{\sqrt{16-2^2}}=\frac{\pi}{6}$$